GATE - 2015 | OS | A computer system implements 8 kilobyte pages and a 32-bit physical address

GATE - 2015 | OS | A computer system implements 8 kilobyte pages and a 32-bit physical address
Posted on 10-02-2022

GATE - 2015 [Operating System]

Question:

A computer system implements 8 kilobyte pages and a 32-bit physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is 24 megabytes, the length of the virtual address supported by the system is ________ bits.  

A

36

B

37

C

38

D

39

    

Solution:

Option (A) is Correct.

Given page size = 8KB = 213B
PAS = 32 bit
∴ No. of frames =PA/Page size = 232 / 213 = 219
Also, it is given that each page table entry contains a valid bit, a dirty bit, 3 permission bits:
= 5 bits reserved
So one Page table entry size is
= 19+5 = 24 bits = 3 bytes
Now, Page table size = No. of entries × Entry size
24 × 220 = No. of entries × 3
No. of entries = 8 × 220 = 2 23
∴ Virtual Address size = No. of entries × Page size = 223 × 213 = 236
∴ Virtual Address Space = 36 bits

Thank You

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